Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $y \neq 0$. $r = \dfrac{4y - 12}{4y^2 - 32y + 60} \times \dfrac{y^2 - 12y + 35}{-3y + 6} $
Solution: First factor out any common factors. $r = \dfrac{4(y - 3)}{4(y^2 - 8y + 15)} \times \dfrac{y^2 - 12y + 35}{-3(y - 2)} $ Then factor the quadratic expressions. $r = \dfrac {4(y - 3)} {4(y - 5)(y - 3)} \times \dfrac {(y - 5)(y - 7)} {-3(y - 2)} $ Then multiply the two numerators and multiply the two denominators. $r = \dfrac {4(y - 3) \times (y - 5)(y - 7) } { 4(y - 5)(y - 3) \times -3(y - 2)} $ $r = \dfrac {4(y - 5)(y - 7)(y - 3)} {-12(y - 5)(y - 3)(y - 2)} $ Notice that $(y - 5)$ and $(y - 3)$ appear in both the numerator and denominator so we can cancel them. $r = \dfrac {4\cancel{(y - 5)}(y - 7)(y - 3)} {-12\cancel{(y - 5)}(y - 3)(y - 2)} $ We are dividing by $y - 5$ , so $y - 5 \neq 0$ Therefore, $y \neq 5$ $r = \dfrac {4\cancel{(y - 5)}(y - 7)\cancel{(y - 3)}} {-12\cancel{(y - 5)}\cancel{(y - 3)}(y - 2)} $ We are dividing by $y - 3$ , so $y - 3 \neq 0$ Therefore, $y \neq 3$ $r = \dfrac {4(y - 7)} {-12(y - 2)} $ $ r = \dfrac{-(y - 7)}{3(y - 2)}; y \neq 5; y \neq 3 $